KE=1/2 m v^2
KE= .5 x 2kg x 15m/s to the 2nd power
KE=225 km/s
(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
Learn more about work done here: brainly.com/question/25573309
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<span>This is best understood with Newtons Third Law of Motion: for every action there is an equal and opposite reaction. That should allow you to see the answer.</span>
Light travels in waves AND in bundles called "photons".
It's hard to imagine something that's a wave and also a bundle.
But it turns out that light behaves like both waves and bundles.
If you design an experiment to detect waves, then it responds to light.
And if you design an experiment to detect 'bundles' or particles, then
that one also responds to light.
